Sets and Probability

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CHAPTER 1 Sets and Probability

In a survey of 200 people that had just returned from a trip to Europe, the following information was gathered.

• 142 visited England

• 95 visited Italy

• 65 visited Germany

• 70 visited both England and Italy

• 50 visited both England and Germany

• 30 visited both Italy and Germany

• 20 visited all three of these countries

How many went to England but not Italy or Germany?

We will learn how to solve puzzles like this in the second section of the chapter when counting the elements in a set is discussed.

1.1 Introduction to Sets

This section discusses operations on sets and the laws governing these set operations. These are fundamental notions that will be used throughout the remainder of this text. In the next two chapters we will see that probability and statistics are based on counting the elements in sets and manipulating set operations. Thus we ﬁrst need to understand clearly the notion of sets and their operations.

✧ The Language of Sets We begin here with some definitions of the language and notation used when working with sets. The most basic definition is “What is a set?” A set is a HISTORICAL NOTE collection of items. These items are referred to as the elements or members of George Boole, 1815–1864 the set. For example, the set containing the numbers 1, 2, and 3 would be written {1,2,3}. Notice that the set is contained in curly brackets. This will help us George Boole was born into a lower-class family in Lincoln, distinguish sets from other mathematical objects. England, and had only a common When all the elements of the set are written out, we refer to this as roster school education. He was largely notation. So the set containing the first 10 letters in the English alphabet would self-taught and managed to { , , , , , , , , , } become an elementary school be written as a b c d e f g h i j in roster notation. If we wanted to refer to teacher. Up to this time any rule of this set without writing all the elements, we could define the set in terms of its algebra such as a(x + y)=ax + ay properties. This is called set-builder notation. So we write was understood to apply only to numbers and magnitudes. Boole { | } developed an “algebra” of sets x x is one of the first 10 letters in the English alphabet where the elements of the sets could be not just numbers but This is read “the set of all x such that x is one of the first 10 letters in the English anything. This then laid down the alphabet”. If we will be using a set more than once in a discussion, it is useful to foundations for a fundamental way of thinking. Bertrand Russell, a define the set with a symbol, usually an uppercase letter. So great mathematician and philosopher of the 20th century, S = {a,b,c,d,e, f ,g,h,i, j} said that the greatest discovery of the 19th century was the nature of { , , , , , , , , , } pure mathematics, which he We can say c is an element of the set a b c d e f g h i j or simply write asserted was discovered by c ∈ S. The symbol ∈ is read “is an element of”. We can also say that the set George Boole. Boole’s pamphlet R = {c} is a subset of our larger set S as every element in the set R is also in the “The Mathematical Analysis of Logic” maintained that the set S. essential character of mathematics lies in its form rather than in its content. Thus mathematics is not Subsets merely the science of If every element of a set A is also an element of another set B, we say that measurement and number but any ⊆ study consisting of symbols and A is a subset of B and write A B.IfA is not a subset of B, we write precise rules of operation. Boole A ⊆ B. founded not only a new algebra of sets but also a formal logic that we will discuss in Chapter L. Thus {1,2,4}⊆{1,2,3,4},but{1,2,3,4} ⊆{1,2,4}. Since every element in A is in A, we can write A ⊆ A. If there is a set B and every element in the set B is also in the set A but B = A, we say that B is a proper subset of A. This is written as B ⊂ A. Note the proper subset symbol ⊂ is lacking the small horizontal line that the subset symbol ⊆ has. The difference is rather like the difference between < and ≤. Some sets have no elements at all. We need some notation for this, simply leaving a blank space will not do!

Empty Set The empty set, written as0or / {}, is the set with no elements.

The empty set can be used to conveniently indicate that an equation has no solution. For example

{x|x is real and x2 = −1} = /0

By the deﬁnition of subset, given any set A, we must have /0 ⊆ A.

EXAMPLE 1 Finding Subsets Find all the subsets of {a,b,c}.

Solution The subsets are /0,{a},{b},{c},{a,b},{a,c},{b,c},{a,b,c} ✦

REMARK: Note that there are 8 subsets and 7 of them are proper subsets. In general, a set with n elements will have 2n subsets. In the next chapter we will learn why this is so.

The empty set is the set with no elements. At the other extreme is the universal set. This set is the set of all elements being considered and is denoted by U. If, for example, we are to take a national survey of voter satisfaction with the president, the universal set is the set of all voters in this country. If the survey is to determine the effects of smoking on pregnant women, the universal set is the set of all pregnant women. The context of the problem under discussion will determine the universal set for that problem. The universal set must contain every element under discussion. A Venn diagram is a way of visualizing sets. The universal set is represented by a rectangle and sets are represented as circles inside the universal set. For Figure 1.1 example, given a universal set U and a set A, Figure 1.1 is a Venn diagram that visualizes the concept that A ⊂ U. Figure 1.1 also visualizes the concept B ⊂ A. The U above the rectangle will be dropped in later diagrams as we will abide by the convention that the rectangle always represents the universal set.

✧ Set Operations The ﬁrst set operation we consider is the complement. The complement of set A are those members of set U that do not belong to A.

Complement Given a universal set U and a set A ⊂ U, the complement of A, written Ac, is the set of all elements that are in U but not in A, that is,

Ac = {x|x ∈ U, x ∈/ A} Figure 1.2 Ac is shaded.

A Venn diagram visualizing Ac is shown in Figure 1.2. Some alternate nota- tions for the complement of a set are A and A¯.

EXAMPLE 2 The Complements of Sets Let U = {1,2,3,4,5,6,7,8,9}, A = {1,3,5,7,9}, B = {1,2,3,4,5}. Find Ac, Bc, Uc,/0c, and (Ac)c in roster notation.

Solution We have Ac = {2,4,6,8} Bc = {6,7,8,9} Uc = /0 /0c = {1,2,3,4,5,6,7,8,9} = U (Ac)c = {2,4,6,8}c = {1,3,5,7,9} = A ✦

Note that in the example above we found Uc = /0and /0c = U. Additionally (Ac)c = A. This can be seen using the Venn diagram in Figure 1.2, since the complement of Ac is all elements in U but not in Ac which is the set A. These three rules are called the Complement Rules.

Complement Rules If U is a universal set, we must always have

Uc = /0, /0c = U

If A is any subset of a universal set U, then

(Ac)c = A

The next set operation is the union of two sets. This set includes the members of both sets A and B. That is, if an element belongs to set A or set B then it belongs to the union of A and B.

Set Union The union of two sets A and B, written A∪B, is the set of all elements that belong to A,ortoB, or to both. Thus

A ∪ B = {x|x ∈ A or x ∈ B or both}

REMARK: This usage of the word “or” is the same as in logic. It is the inclusive “or” where the elements that belong to both sets are part of the union. In English the use of “or” is often the exclusive “or”. That is, if a meal you order at a restaurant comes with a dessert and you are offered cake or pie, you really only get one of the desserts. Choosing one dessert will exclude you from the other. If it was the logical “or” you could have both!

Our convention will be to drop the phrase “or both” but still maintain the same meaning. Note very carefully that this gives a particular deﬁnition to the word “or”. Thus we will normally write

A ∪ B = {x|x ∈ A or x ∈ B}

Figure 1.3 It can be helpful to say that the union of A and B, A ∪ B, is all elements in A A ∪ B is shaded. joined together with all elements in B. A Venn diagram visualizing this is shown in Figure 1.3 with the union shaded.

EXAMPLE 3 The Union of Two Sets Let U = {1,2,3,4,5,6}, A = {1,2,3,4} and B = {1,4,5,6}. Find A ∪ B and A ∪ Ac.

Solution We begin with the ﬁrst set and join to it any elements in the second set that are not already there. Thus

A ∪ B = {1,2,3,4}∪{1,4,5,6} = {1,2,3,4,5,6}

Since Ac = {5,6} we have A ∪ Ac = {1,2,3,4}∪{5,6} = {1,2,3,4,5,6} = U ✦

The second result, A ∪ Ac = U is generally true. From Figure 1.2, we can see that if U is a universal set and A ⊂ U, then

A ∪ Ac = U

Set Intersection The intersection of two sets A and B, written A ∩ B, is the set of all elements that belong to both the set A and to the set B. Thus

A ∩ B = {x|x ∈ A and x ∈ B}

A Venn diagram is shown in Figure 1.4 with the intersection shaded. Figure 1.4 EXAMPLE 4 The Intersection of Two Sets Find

a. {a,b,c,d}∩{a,c,e} b. {a,b}∩{c,d}

Solution a. Only a and c are elements of both of the sets. Thus

{a,b,c,d}∩{a,c,e} = {a,c}

b. The two sets {a,b} and {c,d} have no elements in common. Thus {a,b}∩{c,d} = /0 ✦

The sets {a,b} and {c,d} have no elements in common. These sets are called Figure 1.5 disjoint and can be visualized in Figure 1.5. A and B are disjoint.

Disjoint Sets Two sets A and B are disjoint if they have no elements in common, that is, if A ∩ B = /0.

An examination of Figure 1.2 or referring to the deﬁnition of Ac indicates that for any set A, A and Ac are disjoint. That is,

A ∩ Ac = /0

✧ Additional Laws for Sets There are a number of laws for sets. They are referred to as commutative, associative, distributive, and De Morgan laws. We will consider two of these laws in the following examples. HISTORICAL NOTE EXAMPLE 5 Establishing a De Morgan Law Use a Venn diagram to show Augustus De Morgan, that 1806–1871 (A ∪ B)c = Ac ∩ Bc

It was De Morgan who got George Boole interested in set theory and Solution We ﬁrst consider the right side of this equation. Figure 1.6 shows a c c c c formal logic and then made Venn diagram of A and B and A ∩ B . We then notice from Figure 1.3 that this signiﬁcant advances upon Boole’s is (A ∪ B)c. epochal work. He discovered the De Morgan laws referred to in the last section. Boole and De Morgan are together considered the founders of the algebra of sets and of mathematical logic. De Morgan was a champion of religious and intellectual toleration and on several occasions resigned his professorships in protest of the ✦ abridgments of academic freedom Figure 1.6 of others. EXAMPLE 6 Establishing the Distributive Law for Union Use a Venn diagram to show that A ∪ (B ∩C)=(A ∪ B) ∩ (A ∪C)

Solution Consider ﬁrst the left side of this equation. In Figure 1.7a the sets A, B ∩C, and the union of these two are shown. Now for the right side of the equation refer to Figure 1.7b, where the sets A ∪ B, A ∪C, and the intersection of these two sets are shown. We have the same set in both cases. ✦

Figure 1.7a

Figure 1.7b

We can summarize the laws we have found in the following list.

Laws for Set Operations A ∪ B = B ∪ A Commutative law for union A ∩ B = B ∩ A Commutative law for intersection A ∪ (B ∪C)=(A ∪ B) ∪C Associative law for union A ∩ (B ∩C)=(A ∩ B) ∩C Associative law for intersection A ∪ (B ∩C)=(A ∪ B) ∩ (A ∪C) Distributive law for union A ∩ (B ∪C)=(A ∩ B) ∪ (A ∩C) Distributive law for intersection (A ∪ B)c = Ac ∩ Bc De Morgan law (A ∩ B)c = Ac ∪ Bc De Morgan law

✧ Applications

EXAMPLE 7 Using Set Operations to Write Expressions Let U be the universal set consisting of the set of all students taking classes at the University of Hawaii and B = {x|x is currently taking a business course} E = {x|x is currently taking an English course} M = {x|x is currently taking a math course} Write an expression using set operations and show the region on a Venn diagram for each of the following: a. The set of students at the University of Hawaii taking a course in at least one of the above three ﬁelds.

b. The set of all students at the University of Hawaii taking both an English course and a math course but not a business course.

c. The set of all students at the University of Hawaii taking a course in exactly one of the three ﬁelds above.

Solution a. This is B ∪ E ∪ M. See Figure 1.8a.

b. This can be described as the set of students taking an English course (E) and also (intersection) a math course (M) and also (intersection) not a business course (Bc)or E ∩ M ∩ Bc Figure 1.8a

This is the set of points in the universal set that are in both E and M but not in B and is shown in Figure 1.8b.

c. We describe this set as the set of students taking business but not taking En- glish or math (B ∩ Ec ∩ Mc) together with (union) the set of students taking English but not business or math (E ∩ Bc ∩ Mc) together with (union) the set of students taking math but not business or English (M ∩ Bc ∩ Ec)or Figure 1.8b (B ∩ Ec ∩ Mc) ∪ (Bc ∩ E ∩ Mc) ∪ (Bc ∩ Ec ∩ M)

This is the union of the three sets shown in Figure 1.8c. The ﬁrst, B∩Ec ∩Mc, consists of those points in B that are outside E and also outside M. The second set E ∩ Bc ∩ Mc consists of those points in E that are outside B and M. The third set M ∩ Bc ∩ Ec is the set of points in M that are outside B and E. The union of these three sets is then shown on the right in Figure 1.8c. ✦

REMARK: Figure 1.8c The word only means the same as exactly one. So a student taking only a business course would be written as B ∩ Ec ∩ Mc.

Self-Help Exercises 1.1

1. Let U = {1,2,3,4,5,6,7}, A = {l,2,3,4}, B = a. Corporations in this country that had profits and {3,4,5}, C = {2,3,4,5,6}. Find the following: also paid a dividend last year a. A ∪ B b. A ∩ B b. Corporations in this country that either had prof- c. Ac d. (A ∪ B) ∩C its or paid a dividend last year e. (A ∩ B) ∪C f. Ac ∪ B ∪C c. Corporations in this country that did not have profits last year 2. Let U denote the set of all corporations in this country and P those that made profits during the last year, d. Corporations in this country that had profits, D those that paid a dividend during the last year, paid a dividend, and did not increase their labor and L those that increased their labor force during force last year the last year. Describe the following using the three e. Corporations in this country that had profits or sets P, D, L, and set operations. Show the regions in paid a dividend, and did not increase their labor a Venn diagram. force last year

1.1 Exercises

In Exercises 1 through 4, determine whether the state- 4. a. {x|x(x − 1)=0} = {0,1} ments are true or false. b. {x|x2 + 1 < 0} = /0 1. a. /0 ∈ A b. A ∈ A 5. If A = {u,v,y,z}, determine whether the following 2. a. 0 = /0 b. {x,y}∈{x,y,z} statements are true or false. ∈ ∈/ 3. a. {x|0 < x < −1} = /0 a. w A b. x A b. {x|0 < x < −1} = 0 c. {u,x}∪A d. {y,z,v,u} = A

6. If A = {u,v,y,z}, determine whether the following 16. a. A ∩ B ∩Cc b. B ∩ Ac ∩Cc statements are true or false. a. x ∈/ A b. {u,w} ∈/ A 17. a. Ac ∩ Bc ∩Cc b. A ∩C ∩ Bc c. {x,w} ⊂ A d. /0 ⊂ A 18. a. B ∩C ∩ Ac b. C ∩ Ac ∩ Bc 7. List all the subsets of a. {3}, b. {3,4}. 8. List all the subsets of a. /0, b. {3,4,5}. 19. a. (A ∪ B) ∩Cc b. (A ∩ B)c ∩C

9. Use Venn diagrams to indicate the following. 20. a. A ∪ (B ∩C) b. A ∪ B ∪Cc a. A ⊂ U, B ⊂ U, A ⊂ Bc b. A ⊂ U, B ⊂ U, B ⊂ Ac 21. a. (A ∪ B)c ∩C b. (Ac ∩ B)c ∪C 10. Use Venn diagrams to indicate the following. 22. a. A ∪ (Bc ∩Cc) b. (A ∪ B ∪C)c ∩ A a. A ⊂ U, B ⊂ U, C ⊂ U, C ⊂ (A ∪ B)c b. A ⊂ U, B ⊂ U, C ⊂ U, C ⊂ A ∩ B In Exercises 23 through 30, ﬁnd the indicated sets with U = {1,2,3,4,5,6,7,8,9,10}, For Exercises 11 through 14, indicate where the sets are A = {1,2,3,4,5,6}, located on the ﬁgure below and indicate if the sets found B = {4,5,6,7,8}, C = {5,6,7,8,9,10} in part a and part b are disjoint or not. 23. a. A ∩ B b. A ∪ B

24. a. Ac b. Ac ∩ B

25. a. A ∩ Bc b. Ac ∩ Bc

26. a. Ac ∪ Bc b. (Ac ∪ Bc)c

27. a. A ∩ B ∩C b. (A ∩ B ∩C)c 11. a. A ∩ Bc b. A ∩ B

12. a. Ac ∩ B b. Ac ∩ Bc 28. a. A ∩ (B ∪C) b. A ∩ (Bc ∪C)

13. a. A ∪ Bc b. (A ∪ B)c 29. a. Ac ∩ Bc ∩Cc b. (A ∪ B ∪C)c 14. c ∪ c ( ∩ )c a. A B b. A B 30. a. Ac ∩ Bc ∩C b. Ac ∩ B ∩Cc

For Exercises 15 through 22, indicate where the sets are In Exercises 31 through 34, describe each of the sets in located on the ﬁgure below words.

Let U be the set of all residents of your state and let A = {x|x owns an automobile} H = {x|x owns a house}

31. a. Ac b. A ∪ H c. A ∪ Hc

32. a. Hc b. A ∩ H c. Ac ∩ H

33. a. A ∩ Hc b. Ac ∩ Hc c. Ac ∪ Hc 15. ∩ ∩ ∩ c ∩ c a. A B C b. A B C 34. a. (A ∩ H)c b. (A ∪ H)c c. (Ac ∩ Hc)c

In Exercises 35 through 38, let U, A, and H be as in the b. Major league ball players who have never hit 20 previous four problems, and let homers in a season. P = {x|x owns a piano}, 43. a. New York Yankees or San Francisco Giants who and describe each of the sets in words. have hit 20 homers in a season. 35. a. A ∩ H ∩ P b. A ∪ H ∪ P b. Outfielders for the New York Yankees who have c. (A ∩ H) ∪ P never hit 20 homers in a season. 44. a. Outfielders for the New York Yankees or San ( ∪ ) ∩ ( ∪ ) ∩ c 36. a. A H P b. A H P Francisco Giants. ∩ ∩ c c. A H P b. Outfielders for the New York Yankees who have hit 20 homers in a season. 37. a. (A ∩ H)c ∩ P b. Ac ∩ Hc ∩ Pc c. (A ∪ H)c ∩ P 45. a. Major league outfielders who have hit 20 homers in a season and do not play for the New York Yan- 38. a. (A ∪ H ∪ P)c ∩ A b. (A ∪ H ∪ P)c kees or the San Francisco Giants. c. (A ∩ H ∩ P)c b. Major league outfielders who have never hit 20 homers in a season and do not play for the New York In Exercises 39 through 46, let U be the set of major Yankees or the San Francisco Giants. league baseball players and let 46. a. Major league players who do not play outfield, N = {x|x plays for the New York Yankees} who have hit 20 homers in a season, and do not play S = {x|x plays for the San Francisco Giants} for the New York Yankees or the San Francisco Gi- F = {x|x is an outfielder} ants. H = {x|x has hit 20 homers in one season} Write the set that represents the following descriptions. b. Major league players who play outfield, who have never hit 20 homers in a season, and do not play for the New York Yankees or the San Francisco 39. a. Outfielders for the New York Yankees Giants. b. New York Yankees who have never hit 20 homers in a season In Exercises 47 through 52, let U = {1,2,3,4,5,6,7,8,9,10}, A = {1,2,3,4,5}, 40. a. San Francisco Giants who have hit 20 homers in B = {4,5,6,7}, C = {5,6,7,8,9,10}. a season. Verify that the identities are true for these sets. b. San Francisco Giants who do not play outfield. 47. A ∪ (B ∪C)=(A ∪ B) ∪C 41. a. Major league ball players who play for the New 48. A ∩ (B ∩C)=(A ∩ B) ∩C York Yankees or the San Francisco Giants. b. Major league ball players who play for neither 49. A ∪ (B ∩C)=(A ∪ B) ∩ (A ∪C) the New York Yankees nor the San Francisco Gi- 50. A ∩ (B ∪C)=(A ∩ B) ∪ (A ∩C) ants. 51. (A ∪ B)c = Ac ∩ Bc 42. a. San Francisco Giants who have never hit 20 homers in a season. 52. (A ∩ B)c = Ac ∪ Bc

Solutions to Self-Help Exercises 1.1

1. a. A ∪ B is the elements in A or B. Thus A ∪ B = {1,2,3,4,5}. b. A ∩ B is the elements in both A and B. Thus A ∩ B = {3,4}. c. Ac is the elements not in A (but in U). Thus Ac = {5,6,7}.

d. (A ∪ B) ∩C is those elements in A ∪ B and also in C. From a we have (A ∪ B) ∩C = {1,2,3,4,5}∩{2,3,4,5,6} = {2,3,4,5}

e. (A ∩ B) ∪C is those elements in A ∩ B or in C. Thus from b

(A ∩ B) ∪C = {3,4}∪{2,3,4,5,6} = {2,3,4,5,6}

f. Ac ∪ B ∪C is elements in B,orinC, or not in A. Thus Ac ∪ B ∪C = {2,3,4,5,6,7}

2. a. Corporations in this country that had profits and also paid a dividend last year is represented by P ∩ D. This is regions I and II. b. Corporations in this country that either had profits or paid a dividend last year is represented by P ∪ D. This is regions I, II, III, IV, V, and VI. c. Corporations in this country that did not have profits is represented by Pc. This is regions III, VI, VII, and VIII. d. Corporations in this country that had profits, paid a dividend, and did not increase their labor force last year is represented by P ∩ D ∩ Lc. This is region II. e. Corporations in this country that had profits or paid a dividend, and did not increase their labor force last year is represented by (P∪D)∩Lc. This is regions II, V, and VI.

1.2 The Number of Elements in a Set

In a survey of 120 adults, 55 said they had an egg for breakfast that morn- APPLICATION ing, 40 said they had juice for breakfast, and 70 said they had an egg or Breakfast Survey juice for breakfast. How many had an egg but no juice for breakfast? How many had neither an egg nor juice for breakfast? See Example 1 for the answer.

✧ Counting the Elements of a Set This section shows the relationship between the number of elements in A∪B and the number of elements in A, B, and A ∩ B. This is our ﬁrst counting principle. The examples and exercises in this section give some applications of this. In other applications we will count the number of elements in various sets to ﬁnd probability.

The Notation n(A) If A is a set with a ﬁnite number of elements, we denote the number of elements in A by n(A).

In Figure 1.9 we see the number n(A) written inside the A circle and n(Ac) written outside the set A. This indicates that there are n(A) members in set A and n(Ac) in set Ac. The number of elements in a set is also called the cardinality of the set. There are two results that are rather apparent. First, the empty set /0has no elements n(/0)=0. For the second refer to Figure 1.10 where the two sets A and B are disjoint. Figure 1.9

The Number in the Union of Disjoint Sets If the sets A and B are disjoint, then

n(A ∪ B)=n(A)+n(B)

Figure 1.10 A consequence of the last result is the following. In Figure 1.9, we are given a universal set U and a set A ⊂ U. Then since A ∩ Ac = /0and U = A ∪ Ac,

n(U)=n(A ∪ Ac)=n(A)+n(Ac)

✧ Union Rule for Two Sets Now consider the more general case shown in Figure 1.11. We assume that x is the number in the set A that are not in B, that is, n(A ∩ Bc).Nextwehavez, the number in the set B that are not in A, n(Ac ∩ B). Finally, y is the number in both A and B, n(A ∩ B) and w is the number of elements that are neither in A nor in B, n(Ac ∩ Bc). Then n(A ∪ B)=x + y + z =(x + y)+(y + z) − y = n(A)+n(B) − n(A ∩ B)

Alternatively, we can see that the total n(A)+n(B) counts the number in the Figure 1.11 intersection n(A ∩ B) twice. Thus to obtain the number in the union n(A ∪ B),we must subtract n(A ∩ B) from n(A)+n(B).

The Number in the Union of Two Sets For any ﬁnite sets A and B,

n(A ∪ B)=n(A)+n(B) − n(A ∩ B)

EXAMPLE 1 An Application of Counting In a survey of 120 adults, 55 said they had an egg for breakfast that morning, 40 said they had juice for breakfast, and 70 said they had an egg or juice for breakfast. How many had an egg but no juice for breakfast? How many had neither an egg nor juice for breakfast?

Solution Let U be the universal set of adults surveyed, E the set that had an egg for breakfast, and J the set that had juice for breakfast. A Venn diagram is shown in Figure 1.12a. From the survey, we have that

n(E)=55 n(J)=40, n(E ∪ J)=70

Note that each of these is a sum. That is n(E)=55 = x + y, n(J)=40 = y + z and n(E ∪ J)=70 = x + y + z. Since 120 people are in the universal set, n(U)= Figure 1.12a 120 = x + y + z + w. The number that had an egg and juice for breakfast is given by n(E ∩ J) and is shown as the shaded region in Figure 1.12b. We apply the union rule:

n(E ∩ J)=n(E)+n(J) − n(E ∪ J) = 55 + 40 − 70 = 25

We ﬁrst place the number 25, just found, in the E ∩J area in the Venn diagram in Figure 1.12b. Since the number of people who had eggs (with and without juice) Figure 1.12b is 55, then according to Figure 1.12b,

n(E)=55 = x + 25 x = 30

Similarly, the number who had juice (with and without an egg) is 40. Using Figure 1.12b, n(J)=40 = z + 25 z = 15 These two results are shown in Figure 1.12c. We wish to ﬁnd w = n((E ∪ J)c). This is shown as the shaded region in Figure 1.12c. The unshaded region is E ∪J. We then have that

n(E ∪ J)+n((E ∪ J)c)=n(U) n((E ∪ J)c)=n(U) − n(E ∪ J) w = 120 − 70 w = 50

And so there were 50 people in the surveyed group that had neither an egg nor juice for breakfast. ✦ Figure 1.12c

✧ Counting With Three Sets Many counting problems with sets have two sets in the universal set. We will also study applications with three sets in the universal set. The union rule for three sets is studied in the extensions for this section. In the example below, deductive reasoning is used to solve for the number of elements in each region of the Venn diagram. In cases where this will not solve the problem, systems of linear equations can be used to solve the Venn diagram. This is studied in the Chapter Project found in the Review section.

EXAMPLE 2 European Travels In a survey of 200 people that had just returned from a trip to Europe, the following information was gathered.

• 142 visited England

• 95 visited Italy

• 65 visited Germany

• 70 visited both England and Italy

• 50 visited both England and Germany

• 30 visited both Italy and Germany

• 20 visited all three of these countries

a. How many went to England but not Italy or Germany? b. How many went to exactly one of these three countries? c. How many went to none of these three countries?

Solution Let U be the set of 200 people that were surveyed and let E = {x|x visited England} Figure 1.13a I = {x|x visited Italy} G = {x|x visited Germany}

We ﬁrst note that the last piece of information from the survey indicates that

n(E ∩ I ∩ G)=20

Place this in the Venn diagram shown in Figure 1.13a. Recall that 70 visited both England and Italy, that is, n(E ∩ I)=70. If a is the number that visited England and Italy but not Germany, then, according to Figure 1.13a, 20 + a = n(E ∩ I)=70. Thus a = 50. In the same way, if b is the number that visited England and Germany but not Italy, then 20 + b = n(E ∩ G)=50. Thus b = 30. Also if c is the number that visited Italy and Germany but not England, then 20 + c = n(G ∩ I)=30. Thus c = 10. All of this information is then shown in Figure 1.13b.

Figure 1.13b a. Let x denote the number that visited England but not Italy or Germany. Then, according to Figure 1.13b, 20 + 30 + 50 + x = n(E)=142. Thus x = 42, that is, the number that visited England but not Italy or Germany is 42.

b. Since n(I)=95, the number that visited Italy but not England or Germany is given from Figure 1.13b by 95 − (50 + 20 + 10)=15. Since n(G)=65, the number that visited Germany but not England or Italy is, according to Fig- ure 1.13b, given by 65−(30+20+10)=5. Thus, according to Figure 1.13c, the number who visited just one of the three countries is

42 + 15 + 5 = 62

Figure 1.13c c. There are 200 people in the U and so according to Figure 1.13c, the number that visited none of these three countries is given by 200 − (42 + 15 + 5 + 50 + 30 + 10 + 20)=200 − 172 = 28 ✦

EXAMPLE 3 Pizzas At the end of the day the manager of Blue Baker wanted to know how many pizzas were sold. The only information he had is listed below. Use the information to determine how many pizzas were sold.

• 3 pizzas had mushrooms, pepperoni, and sausage

• 7 pizzas had pepperoni and sausage

• 6 pizzas had mushrooms and sausage but not pepperoni

• 15 pizzas had two or more of these toppings

• 11 pizzas had mushrooms

• 8 pizzas had only pepperoni Figure 1.14a • 24 pizzas had sausage or pepperoni

• 17 pizzas did not have sausage

Solution Begin by drawing a Venn diagram with a circle for pizzas that had mushrooms, a circle for pizzas that had pepperoni, and, pizzas that had sausage. In the center place a 3 since three pizzas had all these toppings. See Figure 1.14a. Since 7 pizzas have pepperoni and sausage, 7 = 3 + III or III = 4. If 6 pizzas had mushrooms and sausage but not pepperoni, then IV = 6. The region for two or more of these toppings is3+II+III+IV=15.Using III = 4 and IV = 6, that gives3+II+4+6=15orII=2.This information is shown in Figure 1.14b. Given that 11 pizzas had mushrooms, V + 2+3+6=11andtherefore V = Figure 1.14b 0. Since 8 pizzas had only pepperoni, VI = 8. With a total of 24 pizzas in the sausage or pepperoni region and knowing that VI = 8, we have 2 + 8 + 6 + 3 + 4 + VII = 24 or VII = 1. Finally, if 17 pizzas did not have sausage then 17 = V + 2 + VI + VIII = 0 + 2 + 8 + VIII. This gives VIII = 7 and our complete diagram is shown in Figure 1.14c. To ﬁnd the total number of pizzas sold, the 8 numbers in the completed Venn diagram are added: 0 + 2 + 8 + 6 + 3 + 4 + 1 + 7 = 31 ✦

Figure 1.14c

Self-Help Exercises 1.2

1. Given that n(A ∪ B)=100, n(A ∩ Bc)=50, and • 40 bought apple juice n(A ∩ B)=20, ﬁnd n(Ac ∩ B). • 19 bought cookies 2. The registrar reported that among 2000 students, • 13 bought broccoli 700 did not register for a math or English course, • 1 bought broccoli, apple juice, and cookies while 400 registered for both of these two courses. • 11 bought cookies and apple juice How many registered for exactly one of these • 2 bought cookies and broccoli but not apple courses? juice 3. One hundred shoppers are interviewed about the • 24 bought only apple juice contents of their bags and the following results are Organize this information in a Venn diagram and found: ﬁnd how many shoppers bought none of these items.

1.2 Exercises

1. If n(A)=100, n(B)=75, and n(A ∩ B)=40, what Applications is n(A ∪ B)? 21. Headache Medicine In a survey of 1200 house- 2. If n(A)=200, n(B)=100, and n(A ∪ B)=250, holds, 950 said they had aspirin in the house, 350 what is n(A ∩ B)? said they had acetaminophen, and 200 said they had 3. If n(A)=100, n(A ∩ B)=20, and n(A ∪ B)=150, both aspirin and acetaminophen. what is n(B)? a. How many in the survey had at least one of the two medications? 4. If n(B)=100, n(A ∪ B)=175, and n(A ∩ B)=40, what is n(A)? b. How many in the survey had aspirin but not acetaminophen? c 5. If n(A)=100 and n(A∩B)=40, what is n(A∩B )? c. How many in the survey had neither aspirin nor acetaminophen? 6. If n(U)=200 and n(A ∪ B)=150, what is n(Ac ∩ Bc)? 22. Newspaper Subscriptions In a survey of 1000 households, 600 said they received the morning pa- 7. If n(A∪B)=500, n(A∩Bc)=200, n(Ac ∩B)=150, per but not the evening paper, 300 said they received what is n(A ∩ B)? both papers, and 100 said they received neither paper. 8. If n(A∩B)=50, n(A∩Bc)=200, n(Ac ∩B)=150, what is n(A ∪ B)? a. How many received the evening paper but not the morning paper? ( ∩ )= ( ∩ ∩ )= 9. If n A B 150 and n A B C 40, what is b. How many received at least one of the papers? n(A ∩ B ∩Cc)? 23. Course Enrollments The registrar reported that 10. If n(A ∩C)=100 and n(A ∩ B ∩C)=60, what is among 1300 students, 700 students did not register n(A ∩ Bc ∩C)? for either a math or English course, 400 registered for an English course, and 300 registered for both 11. If n(A)=200 and n(A ∩ B ∩ C)=40, n(A ∩ B ∩ types of courses. Cc)=20, n(A ∩ Bc ∩C)=50, what is n(A ∩ Bc ∩ a. How many registered for an English course but Cc)? not a math course? 12. If n(B)=200 and n(A ∩ B ∩ C)=40, n(A ∩ B ∩ b. How many registered for a math course? c c c C )=20, n(A ∩ B ∩C)=50, what is n(A ∩ B ∩ 24. Pet Ownership In a survey of 500 people, a pet food c C )? manufacturer found that 200 owned a dog but not a cat, 150 owned a cat but not a dog, and 100 owned For Exercises 13 through 20, let A, B, and C be sets in a neither a dog or cat. universal set U. We are given n(U)=100, n(A)=40, a. How many owned both a cat and a dog? n(B)=37, n(C)=35, n(A ∩ B)=25, n(A ∩C)=22, b. How many owned a dog? ( ∩ )= ( ∩ ∩ c)= n B C 24, and n A B C 10. Find the follow- 25. Fast Food A survey by a fast-food chain of 1000 ing values. adults found that in the past month 500 had been to Burger King, 700 to McDonald’s, 400 to Wendy’s, ( ∩ ∩ ) ( c ∩ ∩ ) 13. n A B C 14. n A B C 300 to Burger King and McDonald’s, 250 to Mc- Donald’s and Wendy’s, 220 to Burger King and ( ∩ c ∩ ) ( ∩ c ∩ c) 15. n A B C 16. n A B C Wendy’s, and 100 to all three. How many went to ( c ∩ ∩ c) ( c ∩ c ∩ ) 17. n A B C 18. n A B C a. Wendy’s but not the other two? b. only one of them? 19. n(A ∪ B ∪C) 20. n((A ∪ B ∪C))c c. none of these three?

26. Investments A survey of 600 adults over age 50 29. Magazines In a survey of 250 business executives, found that 200 owned some stocks and real estate 40 said they did not read Money, Fortune, or Busi- but no bonds, 220 owned some real estate and bonds ness Week, while 120 said they read exactly one but no stock, 60 owned real estate but no stocks or of these three and 60 said they read exactly two of bonds, and 130 owned both stocks and bonds. How them. How many read all three? many owned none of the three? 30. Sales A furniture store held a sale that attracted 100 27. Entertainment A survey of 500 adults found that people to the store. Of these, 57 did not buy any- 190 played golf, 200 skied, 95 played tennis, 100 thing, 9 bought both a sofa and love seat, 8 bought played golf but did not ski or play tennis, 120 skied both a sofa and chair, 7 bought both a love seat but did not play golf or tennis, 30 played golf and and chair. There were 24 sofas, 18 love seats, and skied but did not play tennis, and 40 did all three. 20 chairs sold. How many people bought all three a. How many played golf and tennis but did not items? ski? b. How many played tennis but did not play golf or Extensions ski? c. How many participated in at least one of the 31. Use a Venn diagram to show that three sports? 28. Transportation A survey of 600 adults found that n(A ∪ B ∪C)=n(A)+n(B)+n(C) during the last year, 100 traveled by plane but not −n(A ∩ B) − n(A ∩C) by train, 150 traveled by train but not by plane, 120 −n(B ∩C)+n(A ∩ B ∩C) traveled by bus but not by train or plane, 100 traveled by both bus and plane, 40 traveled by all three, 32. Give a proof of the formula in Exercise 31. Hint: and 360 traveled by plane or train. How many did Set B ∪C = D and use union rule on n(A ∪ D). Now use the not travel by any of these three modes of transporta- union rule two more times, recalling from the last section that tion? A ∩ (B ∪C)=(A ∩ B) ∪ (A ∩C).

Solutions to Self-Help Exercises 1.2

1. The accompanying Venn diagram indicates that

n(A ∩ Bc)=50, n(A ∩ B)=20, z = n(Ac ∩ B)

Then, according to the diagram,

50 + 20 + z = n(A ∪ B)=100

Thus z = 30.

2. The number of students that registered for exactly one of the courses is the number that registered for math but not English, x = n(M ∩ Ec), plus the number that registered for English but not math, z = n(Mc ∩ E). Then, according to the accompanying Venn diagram, x+z+400+700 = 2000. Thus x+z = 900. That is, 900 students registered for exactly one math or English course.

3. Let A be the set of shoppers who bought apple juice, B the set of shoppers who bought broccoli, and C the set of shoppers who bought cookies. This is shown in the ﬁrst ﬁgure below. Since one shopper bought all three items, a 1 is placed in region I. Twenty-four shoppers bought only apple juice and this is region V.

Given 2 shoppers bought cookies and broccoli but not apple juice, a 2 is placed in region III. This is shown in the next figure below. The statement “11 bought cookies and apple juice” includes those who bought broccoli and those who did not. We now know that one person bought all 3 items, so 11 − 1 = 10 people bought cookies and apple juice but not broccoli. A 10 is placed in region IV. NowI+II+IV+V=40aswearetold “40 bought apple juice.” With the 10 in region IV we know 3 of the 4 values for set A and we can solve for region II: 40 = 24 + 10 + 1 + II gives II = 5. Place this in the Venn diagram as shown in the third figure below. Examining the figure, we can use the total of 13 in the broccoli circle to solve for VI: 18 = 5 + 1 + 2 + VI gives VI = 10. The total of 19 in the cookies circle lets us solve for VII: 10 +1+2+VII=13givesVII=0.Theverylast piece of information is that there were 100 shoppers. To solve for VIII we have 100 = 24 + 5 + 10 + 10 + 1 + 2 + 0 + VIII or VIII = 48. That is, 48 shoppers bought none of these items. The completed diagram is the final figure below.

1.3 Sample Spaces and Events

Many people have a good idea of the basics of probability. That is, if a fair coin is flipped, you have an equal chance of a head or a tail showing. However, as we proceed to study more advanced concepts in probability we need some formal definitions that will both agree with our intuitive understanding of probability and allow us to go deeper into topics such as conditional probability. This will tie closely to work we have done learning about sets. ✧ The Language of Probability We begin the preliminaries by stating some definitions. It is very important to have a clear and precise language to discuss probability so pay close attention to the exact meanings of the terms below.

Experiments and Outcomes An experiment is an activity that has observable results. An outcome is the result of the experiment.

The following are some examples of experiments. Flip a coin and observe whether it falls “heads” or “tails.” Throw a die (a small cube marked on each face with from one to six dots1) and observe the number of dots on the top face. Select a transistor from a bin and observe whether or not it is defective.

The following are some additional terms that are needed.

Sample Spaces and Trials A sample space of an experiment is the set of all possible outcomes of the experiment. Each repetition of an experiment is called a trial.

For the experiment of throwing a die and observing the number of dots on the top face the sample space is the set

S = {1,2,3,4,5,6}

In the experiment of ﬂipping a coin and observing whether it falls heads or tails, the sample space is S = {heads,tails} or simply S = {H,T}.

EXAMPLE 1 Determining the Sample Space An experiment consists of not- ing whether the price of the stock of the Ford Corporation rose, fell, or remained unchanged on the most recent day of trading. What is the sample space for this experiment?

Solution There are three possible outcomes depending on whether the price rose, fell, or remained unchanged. Thus the sample space S is = { , , } S rose fell unchanged ✦ . EXAMPLE 2 Determining the Sample Space Two dice, identical except that one is green and the other is red, are tossed and the number of dots on the top face of each is observed. What is the sample space for this experiment?

Solution Each die can take on its six different values with the other die also taking on all of its six different values. We can express the outcomes as order pairs. For example, (2, 3) will mean 2 dots on the top face of the green die and 3 dots on the top face of the red die. The sample space S is below. A more colorful version is shown in Figure 1.15.

Figure 1.15 S = {(1,1),(1,2),(1,3),(1,4),(1,5),(1,6), (2,1),(2,2),(2,3),(2,4),(2,5),(2,6), (3,1),(3,2),(3,3),(3,4),(3,5),(3,6), (4,1),(4,2),(4,3),(4,4),(4,5),(4,6), (5,1),(5,2),(5,3),(5,4),(5,5),(5,6), (6,1),(6,2),(6,3),(6,4),(6,5),(6,6)} ✦

1There are also four-sided die, eight-sided die, and so on. However, the six-sided die is the most common and six-sided should be assumed when we refer to a die, unless otherwise speciﬁed.

If the experiment of tossing 2 dice consists of just observing the total number of dots on the top faces of the two dice, then the sample space would be

S = {2,3,4,5,6,7,8,9,10,11,12}

In short, the sample space depends on the precise statement of the experiment.

EXAMPLE 3 Determining the Sample Space A coin is ﬂipped twice to observe whether heads or tails shows; order is important. What is the sample space for this experiment?

Solution The sample space S consists of the 4 outcomes S = {(H,H),(H,T),(T,H),(T,T)} ✦

✧ Tree Diagrams In Example 3 we completed a task (flipped a coin) and then completed another task (flipped the coin again). In these cases, the experiment can be diagrammed Figure 1.16 with a tree. The tree diagram for Example 3 is shown in Figure 1.16. We see we have a first set of branches representing the first flip of the coin. From there we flip the coin again and have a second set of branches. Then trace along each branch to find the outcomes of the experiment. If the coin is tossed a third time, there will be eight outcomes.

EXAMPLE 4 Determining the Sample Space A die is rolled. If the die shows a 1 or a 6, a coin is tossed. What is the sample space for this experiment?

Solution Figure 1.17 shows the possibilities. We then have

S = {(1,H),(1,T),2,3,4,5,(6,H),(6,T)} ✦

Figure 1.17 ✧ Events We start this subsection with the following deﬁnition of an event.

Events and Elementary Events Given a sample space S for an experiment, an event is any subset E of S. An elementary (or simple) event is an event with a single outcome.

EXAMPLE 5 Finding Events Using the sample space from Example 3 ﬁnd the events: “At least one head comes up” and “Exactly two tails come up.” Are either events elementary events?

Solution “At least one head comes up” = {(H,H),(H,T),(T,H)} “Exactly two tails come up” = {(T,T)} The second event, “Exactly two tails come up” has only one outcome and so it is an elementary event. ✦

We can use our set language for union, intersection, and complement to describe events.

Union of Two Events If E and F are two events, then E ∪ F is the union of the two events and consists of the set of outcomes that are in E or F.

Thus the event E ∪ F is the event that “E or F occurs.” Refer to Figure 1.18 where the event E ∪ F is the shaded region on the Venn diagram.

Figure 1.18

Intersection of Two Events If E and F are two events, then E ∩ F is the intersection of the two events and consists of the set of outcomes that are in both E and F.

Thus the event E ∩F is the event that “E and F both occur.” Refer to Figure 1.18 where the event E ∩ F is the region where E and F overlap.

Complement of an Event If E is an event, then Ec is the complement of E and consists of the set of outcomes that are not in E.

Thus the event Ec is the event that “E does not occur.”

EXAMPLE 6 Determining Union, Intersection, and Complement Consider the sample space given in Example 2. Let E consist of those outcomes for which the number of dots on the top faces of both dice is 2 or 4. Let F be the event that the sum of the number of dots on the top faces of the two dice is 6. Let G be the event that the sum of the number of dots on the top faces of the two dice is less than 11. a. List the elements of E and F. b. Find E ∪ F. c. Find E ∩ F. d. Find Gc.

Solution a. E = {(2,2),(2,4),(4,2),(4,4)} and F = {(1,5),(2,4),(3,3),(4,2),(5,1)} b. E ∪ F = {(2,2),(2,4),(4,2),(4,4),(1,5),(3,3),(5,1)} c. E ∩ F = {(2,4),(4,2)} d. Gc = {(5,6),(6,5),(6,6)} ✦

If S is a sample space, /0 ⊆ S, and thus /0is an event. We call the event /0the impossible event since the event /0means that no outcome has occurred, whereas, in any experiment some outcome must occur.

The Impossible Event The empty set, /0,is called the impossible event.

For example, if H is the event that a head shows on ﬂipping a coin and T is the event that a tail shows, then H ∩ T = /0. The event H ∩ T means that both heads and tails shows, which is impossible.

Since S ⊆ S, S is itself an event. We call S the certainty event since any outcome of the experiment must be in S. For example, if a fair coin is ﬂipped, the event H ∪ T is certain since a head or tail must occur.

The Certainty Event Let S be a sample space. The event S is called the certainty event.

We also have the following deﬁnition for mutually exclusive events. See Figure 1.19.

Mutually Exclusive Events Two events E and F are said to be mutually exclusive if the sets are disjoint. That is, E ∩ F = /0 Figure 1.19

Standard Deck of 52 Playing Cards A standard deck of 52 playing cards has four 13-card suits: clubs ♣, diamonds ♦, hearts ♥, and spades ♠. The diamonds and hearts are red, while the clubs and spades are black. Each 13-card suit contains cards numbered from 2 to 10, a jack, a queen, a king, and an ace. The jack, queen, king, and ace can be considered respectively as number 11, 12, 13, and 14. In poker the ace can be either a 14 or a 1. See Figure 1.20.

Figure 1.20

EXAMPLE 7 Determining if Sets Are Mutually Exclusive Let a card be chosen from a standard deck of 52 cards. Let E be the event consisting of drawing a 3. Let F be the event of drawing a heart. Let G be the event of drawing a Jack. Are E and F mutually exclusive? Are E and G?

Solution Since E ∩ F is the event that the card is a 3 and is a heart,

E ∩ F = {3♥} = /0

and so these events are not mutually exclusive. The event E ∩ G is the event that the card is a 3 and a jack, so E ∩ G = /0 therefore E and G are mutually exclusive. ✦

✧ Continuous Sample Spaces In all the previous examples we were able to list each outcome in the sample space, even if the list is rather long. But consider the outcomes of an experiment where the time spent running a race is measured. Depending on how the time is measured, an outcome could be 36 seconds or 36.0032 seconds. The values of the outcomes are not restricted to whole numbers and so the sample space must be described, rather than listed. In the case of the race we could say S = {t|t ≥ 0,t in seconds}. Then the event E that a person takes less than 35 seconds to run the race would be written E = {t|t < 35 seconds}.

EXAMPLE 8 Weighing Oranges At a farmer’s market there is a display of fresh oranges. The oranges are carefully weighed. What is a sample space for this experiment? Describe the event that a orange weighs 100 grams or more. Describe the event that a orange weighs between 200 and 250 grams.

Solution Since the weight of the orange can be any positive number,

S = {w|w > 0,w in grams}

Note that w = 0 is not included as if the weight was zero there would be no orange! The event that an orange weighs 100 grams or more is

E = {w|w ≥ 100,w in grams}

Here note that we use ≥, not > as the value of exactly 100 grams needs to be included. The event that the orange weighs between 200 and 250 grams is

F = {w|200 < w < 250,w in grams}

where strict inequalities are used as the weight is between those values. ✦

Self-Help Exercises 1.3

1. Two tetrahedrons (4 sided), each with equal sides a. List the elements of E and F. numbered from 1 to 4, are identical except that one b. Find E ∩ F. is red and the other green. If the two tetrahedrons c. Find E ∪ F. are tossed and the number on the bottom face of c each is observed, what is the sample space for this d. Find G . experiment? 3. A hospital carefully measures the length of every 2. Consider the sample space given in the previous ex- baby born. What is a sample space for this experi- ercise. Let E consist of those outcomes for which ment? Describe the events both (tetrahedron) dice show an odd number. Let a. the baby is longer than 22 inches. F be the event that the sum of the two numbers on these dice is 5. Let G be the event that the sum of b. the baby is 20 inches or shorter. the two numbers is less than 7. c. the baby is between 19.5 and 21 inches long.

1.3 Exercises

1. Let S = {a,b,c} be a sample space. Find all the consists of selecting a ball from the urn and observ- events. ing its color. What is a sample space for this experiment? Indicate the outcomes in the event “the ball 2. Let the sample space be S = {a,b,c,d}. How many is not white.” events are there? 8. For the urn in Exercise 7, an experiment consists of 3. A coin is ﬂipped three times, and heads or tails is selecting 2 balls in succession without replacement observed after each ﬂip. What is the sample space? and observing the color of each of the balls. What Indicate the outcomes in the event “at least 2 heads is the sample space of this experiment? Indicate the are observed.” outcomes of the event “no ball is white.”

4. A coin is flipped, and it is noted whether heads or 9. Ann, Bubba, Carlos, David, and Elvira are up for tails show. A die is tossed, and the number on the promotion. Their boss must select three people from top face is noted. What is the sample space of this this group of five to be promoted. What is the experiment? sample space? Indicate the outcomes of the event “Bubba is selected.” 5. A coin is flipped three times. If heads show, one is written down. If tails show, zero is written down. 10. A restaurant offers six side dishes: rice, macaroni, What is the sample space for this experiment? Indi- potatoes, corn, broccoli, and carrots. A customer cate the outcomes if “one is observed at least twice.” must select two different side dishes for his dinner. What is the sample space? List the outcomes of the 6. Two tetrahedrons (4 sided), each with equal sides event “Corn is selected.” numbered from 1 to 4, are identical except that one is red and the other green. If the two tetrahedrons 11. An experiment consists of selecting a digit from are tossed and the number on the bottom face of the number 112964333 and observing it. What is each is observed, indicate the outcomes in the event a sample space for this experiment? Indicate the “the sum of the numbers is 4.” outcomes in the event that “an even digit.”

7. An urn holds 10 identical balls except that 1 is 12. An experiment consists of selecting a letter from the white, 4 are black, and 5 are red. An experiment word CONNECTICUT and observing it. What is a

sample space for this experiment? Indicate the out- d. Find and describe the sets E ∪ G, E ∩ G, Gc, comes of the event “a vowel is selected.” Ec ∩ G, and (E ∪ G)c. e. Find all pairs of sets listed in part (d) that are 13. An inspector selects 10 transistors from the produc- mutually exclusive. tion line and notes how many are defective. 16. Let E be the event that the life of a certain light bulb a. Determine the sample space. is at least 100 hours and F that the life is at most b. Find the outcomes in the set corresponding to 200 hours. Describe the sets: the event E “at least 6 are defective.” a. E ∩ F b. Fc c. Ec ∩ F d. (E ∪ F)c c. Find the outcomes in the set corresponding to the event F “at most 4 are defective.” 17. Let E be the event that a pencil is 10 cm or longer d. Find the sets E ∪ F, E ∩ F, Ec, E ∩ Fc, Ec ∩ Fc. and F the event that the pencil is less than 25 cm. Describe the sets: e. Find all pairs of sets among the nonempty ones ∩ c ∩ c ( ∪ )c listed in part (d) that are mutually exclusive. a. E F b. E c. E F d. E F 14. A survey indicates first whether a person is in the In Exercises 18 through 23, S is a sample space and E, ∩ ∪ c lower income group (L), middle income group (M), F, and G are three events. Use the symbols , , and or upper income group (U), and second which of to describe the given events. these groups the father of the person is in. 18. F but not E 19. E but not F a. Determine the sample space using the letters L, M, and U. 20. Not F or not E 21. Not F and not E b. Find the outcomes in the set corresponding to 22. Not F, nor E, nor G the event E “the person is in the lower income group.” 23. E and F but not G c. Find the outcomes in the set corresponding to 24. Let S be a sample space consisting of all the integers the event F “the person is in the higher income from 1 to 20 inclusive, E the first 10 of these, and group.” F the last 5 of these. Find E ∩ F, Ec ∩ F, (E ∪ F)c, c c d. Find the sets E ∪ F, E ∩ F, Ec, E ∩ Fc, Ec ∩ Fc. and E ∩ F . e. Find all pairs of sets listed in part (d) that are 25. Let S be the 26 letters of the alphabet, E be the vow- mutually exclusive. els {a,e,i,o,u}, F the remaining 21 letters, and G 15. A corporate president decides that for each of the the first 5 letters of the alphabet. Find the events ∪ ∪ c ∪ c ∪ c ∩ ∩ ∪ c ∪ next three fiscal years success (S) will be declared if E F G, E F G , E F G, and E F G. the earnings per share of the company go up at least 26. A bowl contains a penny, a nickel, and a dime. 10% that year and failure (F) will occur is less than A single coin is chosen at random from the bowl. 10%. What is the sample space for this experiment? List a. Determine the sample space using the letters S the outcomes in the event that a penny or a nickel is and F. chosen. b. Find the outcomes in the set corresponding to 27. A cup contains four marbles. One red, one blue, one the event E “at least 2 of the next 3 years is a green, and one yellow. A single marble is drawn at success.” random from the cup. What is the sample space for c. Find the outcomes in the set corresponding to this experiment? List the outcomes in the event that the event G “the first year is a success.” a blue or a green marble is chosen.

Solutions to Self-Help Exercises 1.3

1. Consider the outcomes as ordered pairs, with the number on the bottom of the red one the ﬁrst number and the number on the bottom of the white one the second number. The sample space is

S = { (1,1),(1,2),(1,3),(1,4), (2,1),(2,2),(2,3),(2,4), (3,1),(3,2),(3,3),(3,4), (4,1),(4,2),(4,3),(4,4)}

2. a. E = {(1,1),(1,3),(3,1),(3,3)}, and F = {(1,4),(2,3),(3,2),(4,1)} b. E ∩ F = /0 c. E ∪ F = {(1,1),(1,3),(3,1),(3,3),(1,4),(2,3),(3,2),(4,1)} d. Gc = {(3,4),(4,3),(4,4)}

3. Since the baby can be any length greater than zero, the sample space is

S = {x|x > 0,x in inches}

a. E = {x|x > 22,x in inches} b. F = {x|x ≤ 20,x in inches} c. G = {x|19.5 < x < 21,x in inches}

1.4 Basics of Probability

✧ Introduction to Probability We first consider sample spaces for which the outcomes (elementary events) are equally likely. For example, a head or tail is equally likely to come up on a flip of a fair coin. Any of the six numbers on a fair die is equally likely to come up on a roll. We will refer to a sample space S whose individual elementary events are equally likely as a uniform sample space. We then give the following definition of the probability of any event in a uniform sample space.

Probability of an Event in a Uniform Sample Space If S is a ﬁnite uniform sample space and E is any event, then the probability of E, P(E), is given by

Number of elements in E n(E) P(E)= = Number of elements in S n(S)